Câu hỏi của Mộc Lung Hoa

Question

giải pt bậc 2 một ẩn:

a,$\left(1+\sqrt{2}\right)$x^2 $-x-\sqrt{2}$=0

Nhã Doanh · Accepted Answer

$\left(1+\sqrt{2}\right)x^2-x-\sqrt{2}=0$

$\Leftrightarrow\left(1+\sqrt{2}\right)x^2-x-\sqrt{2}x+\sqrt{2}x-\sqrt{2}=0$

$\Leftrightarrow\left(1+\sqrt{2}\right)x^2-x\left(1+\sqrt{2}\right)+\sqrt{2}\left(x-1\right)=0$

$\Leftrightarrow\left(1+\sqrt{2}\right)x\left(x-1\right)+\sqrt{2}\left(x-1\right)=0$

$\Leftrightarrow\left(x-1\right)\left(x+\sqrt{2}x+\sqrt{2}\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=1\x=-2+\sqrt{2}\end{matrix}\right.$Câu trả lời: $\left(1+\sqrt{2}\right)x^2-x-\sqrt{2}=0$

$\Leftrightarrow\left(1+\sqrt{2}\right)x^2-x-\sqrt{2}x+\sqrt{2}x-\sqrt{2}=0$

$\Leftrightarrow\left(1+\sqrt{2}\right)x^2-x\left(1+\sqrt{2}\right)+\sqrt{2}\left(x-1\right)=0$

$\Leftrightarrow\left(1+\sqrt{2}\right)x\left(x-1\right)+\sqrt{2}\left(x-1\right)=0$

$\Leftrightarrow\left(x-1\right)\left(x+\sqrt{2}x+\sqrt{2}\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=1\x=-2+\sqrt{2}\end{matrix}\right.$

Bài 1: Phương trình bậc nhất hai ẩn